Bài 1: tìm x a) (3x-2)²=81 b)(3 x+1)³=-27 Bài 2: giá trị nhỏ nhất của: A= (x-4)²+1/2 B= (3x+1)²-6 C= (4x-5)²-5/4

Đáp án:

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Bài `1`

`a,`

`(3x – 2)^2 = 81`

`↔` \(\left[ \begin{array}{l}(3x – 2)^2 = 9^2\\(3x-2)^2=(-9)^2\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}3x-2=9\\3x-2=-9\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}3x=9+2\\3x=-9+2\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}3x=11\\3x=-7\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=11÷3\\x=-7÷3\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=\dfrac{11}{3}\\x=\dfrac{-7}{3}\end{array} \right.\) 

Vậy `x=11/3` hoặc `x=(-7)/3`

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`b,`

`(3x+1)^3=27`

`↔ (3x+1)^3=3^3`

`↔3x+1=3`

`↔3x=3-1`

`↔3x=2`

`↔x=2÷3`

`↔x=2/3`

Vậy `x=2/3`

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Bài `2`

`A = (x-4)^2+1/2`

Vì `(x-4)^2≥ 0∀x`

`↔ (x-4)^2+1/2≥1/2`

`↔A≥1/2`

`↔ min A = 1/2`

Dấu “`=`” xảy ra khi :

`↔(x-4)^2=0`

`↔x-4=0`

`↔x=4`

Vậy `min A = 1/2 ↔ x=4`

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`B = (3x+1)^2-6`

Vì `(3x+1)^2≥0∀x`

`↔ (3x+1)^2-6≥ -6`

`↔ B ≥ -6`

`↔ min B = -6`

Dấu “`=`” xảy ra khi :

`↔ (3x+1)^2=0`

`↔3x+1=0`

`↔3x=-1`

`↔x=(-1)/3`

Vậy `min B = -6 ↔x=(-1)/3`

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`C = (4x-5)^2-5/4`

Vì `(4x-5)^2≥0∀x`

`↔ (4x-5)^2-5/4 ≥ (-5)/4`

`↔ C ≥ (-5)/4`

`↔ min C = (-5)/4`

Dấu “`=`” xảy ra khi :

`↔(4x-5)^2=0`

`↔4x-5=0`

`↔4x=5`

`↔x=5/4`

Vậy `min C = (-5)/4 ↔ x=5/4`

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